Question
I've inherited $400,000 and I'm considering taking monthly withdrawals for 16 years at a 4% annual interest rate. Can you find the monthly withdrawal amount for me?
Answers
GPT-4o mini
To calculate the monthly withdrawal amount from an inherited amount of $400,000 over 16 years, with an annual interest rate of 4%, we can use the formula for the annuity withdrawal:
\[
M = \frac{P \cdot r}{1 - (1 + r)^{-n}}
\]
Where:
- \(M\) is the monthly withdrawal amount,
- \(P\) is the principal (initial amount, which is $400,000),
- \(r\) is the monthly interest rate (annual rate divided by 12), and
- \(n\) is the total number of withdrawals (total months).
First, let's convert the annual interest rate of 4% to a monthly interest rate:
\[
r = \frac{0.04}{12} = \frac{0.04}{12} = 0.0033333
\]
Now, let's calculate the total number of months for 16 years:
\[
n = 16 \cdot 12 = 192
\]
Now we can plug the values into the formula:
\[
M = \frac{400000 \cdot 0.0033333}{1 - (1 + 0.0033333)^{-192}}
\]
Calculating the denominator:
1. Calculate \((1 + r)^{-n}\):
\[
(1 + 0.0033333)^{-192} \approx 0.470673
\]
2. Now calculate \(1 - (1 + r)^{-n}\):
\[
1 - 0.470673 \approx 0.529327
\]
Now substitute this back into the equation for \(M\):
\[
M = \frac{400000 \cdot 0.0033333}{0.529327} \approx \frac{1333.32}{0.529327} \approx 2517.73
\]
So, the monthly withdrawal amount would be approximately **$2,517.73**.
\[
M = \frac{P \cdot r}{1 - (1 + r)^{-n}}
\]
Where:
- \(M\) is the monthly withdrawal amount,
- \(P\) is the principal (initial amount, which is $400,000),
- \(r\) is the monthly interest rate (annual rate divided by 12), and
- \(n\) is the total number of withdrawals (total months).
First, let's convert the annual interest rate of 4% to a monthly interest rate:
\[
r = \frac{0.04}{12} = \frac{0.04}{12} = 0.0033333
\]
Now, let's calculate the total number of months for 16 years:
\[
n = 16 \cdot 12 = 192
\]
Now we can plug the values into the formula:
\[
M = \frac{400000 \cdot 0.0033333}{1 - (1 + 0.0033333)^{-192}}
\]
Calculating the denominator:
1. Calculate \((1 + r)^{-n}\):
\[
(1 + 0.0033333)^{-192} \approx 0.470673
\]
2. Now calculate \(1 - (1 + r)^{-n}\):
\[
1 - 0.470673 \approx 0.529327
\]
Now substitute this back into the equation for \(M\):
\[
M = \frac{400000 \cdot 0.0033333}{0.529327} \approx \frac{1333.32}{0.529327} \approx 2517.73
\]
So, the monthly withdrawal amount would be approximately **$2,517.73**.